What is the probability of getting STRAIGHT FLUSH in a $13$-card poker game?

Here is my attempt:

A straight flush is five cards in sequence and of the same suit, but not ace king queen jack ten.

The required probability is $$\dfrac{9 \cdot {4 \choose 1}}{{52 \choose 13}} = \dfrac{36}{635013559600} = \dfrac{9}{158753389900} \approx 5.66917028081678777\ldots \cdot{10}^{-11}.$$

My questions are:

(1) Is this probability computation correct?

(2) If my computation is not correct, where is/are the error(s) and what hint can you give towards rectifying that error(s)?

  • $\begingroup$ As in the question about $4$ of a kind, the count is very much too small. It does not take into account the other $8$ cards. Also, do two straight flush count? An Inclusion/Exclusion argument will work. $\endgroup$ – André Nicolas Apr 16 '16 at 1:55
  • $\begingroup$ @AndréNicolas, yes - two straight flushes will count. Will an Inclusion/Exclusion argument similar to that in the question about $4$ of a kind, work? $\endgroup$ – Jose Arnaldo Bebita-Dris Apr 16 '16 at 2:16
  • 1
    $\begingroup$ There are $13 \choose 5$ ways to obtain a given straight flush among the 13 card hand, so multiply your result by that $\endgroup$ – frogfanitw Apr 16 '16 at 2:20
  • $\begingroup$ btw, who plays 13-card poker? $\endgroup$ – frogfanitw Apr 16 '16 at 2:23
  • $\begingroup$ @frogfanitw, 13-card poker is also known as Chinese poker. $\endgroup$ – Jose Arnaldo Bebita-Dris Apr 16 '16 at 2:24

Your are missing the extra cards
It is 52 - 6 as you cannot have the card above of it makes a higher straight

This is for one

$$\frac {\binom{9}{1} \binom{4}{1} \binom{46}{8} } { \binom{52}{13} } = 0.01479 \approx 1 / 67.6$$

I think this is 2

$$\frac {\binom{9}{1} \binom{4}{2} \binom{40}{3} } { \binom{52}{13} } = 0.00000084 \approx 1 / 1190234$$


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